Selina Solutions Concise Mathematics Class 6 Chapter 26 Triangles provides students with a clear idea about the basic concepts discussed in this chapter. The solutions are created by expert teachers after conducting vast research on each topic, according to the grasping capacity of students. For further details, students can make use of Selina Solutions Concise Mathematics Class 6 Chapter 26 Triangles free PDF, from the links mentioned here.
Chapter 26 gives a clear idea about various types of triangles relying on the angles and length of sides. Students who aim to improve problem solving analytical thinking skills are suggested to practice solutions on a regular basis. This also helps them to clear their doubts immediately, about the concepts covered.
Selina Solutions Concise Mathematics Class 6 Chapter 26: Triangles Download PDF
Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 26: Triangles
Access Selina Solutions Concise Mathematics Class 6 Chapter 26: Triangles
Exercise 26(A)
1. In each of the following, find the marked unknown angles:
Solution:
(i) We know that,
Sum of all angles of triangle = 1800
Therefore,
700 + 720 + z = 1800
1420 + z = 1800
z = 1800 – 1420
We get,
z = 380
(ii) We know that,
Sum of all angles of a triangle = 1800
First triangle
500 + 800 + b = 1800
1300 + b = 1800
b = 1800 – 1300
We get,
b = 500
Second triangle
400 + 450 + a = 1800
850 + a = 1800
a = 1800 – 850
We get,
a = 950
(iii) 600 + 450 + 200 + x = 1800
1250 + x = 1800
x = 1800 – 1250
We get,
x = 550
2. Can a triangle together have the following angles?
(i) 550, 550 and 800
(ii) 330, 740 and 730
(iii) 850, 950 and 220
Solution:
(i) Sum of all angles of a triangle = 1800
Here,
550 + 550 + 800 = 1800
We get,
1900 ≠1800
Therefore, it cannot form a triangle
(ii) 330 + 740 + 730 = 1800
We get,
1800 = 1800
Therefore, it form a triangle
(iii) 850 + 950 + 220 = 1800
We get,
2020 ≠1800
Therefore, it cannot form a triangle
3. Find x, if the angles of a triangle are:
(i) x0, x0, x0
(ii) x0, 2x0, 2x0
(iii) 2x0, 4x0, 6x0
Solution:
We know that,
The sum of all the angles in a triangle is 1800
So,
x0 + x0 + x0 = 1800
3x = 1800
x = 1800 / 3
We get,
x = 600
The value of x = 600
(ii) We know that,
The sum of all the angles in a triangle is 1800
So,
x + 2x + 2x = 1800
5x = 1800
x = 1800 / 5
We get,
x = 360
Therefore, the value of x = 360
(iii) We know that,
The sum of all the angles in a triangle is 1800
So,
2x + 4x + 6x = 1800
12x = 1800
x = 1800 / 12
We get,
x = 150
Therefore, the value of x = 150
4. One angle of a right-angled triangle is 700. Find the other acute angle.
Solution:
We know that,
Sum of all the angles in a triangle = 1800
Let us consider the acute angle as x
Hence,
x + 900 + 700 = 1800
x + 1600 = 1800
x = 1800 – 1600
We get,
x = 200
Therefore, the acute angle is 200
5. In △ABC, ∠A = ∠B = 620; find ∠C
Solution:
Given
∠A = ∠B = 620
So,
∠A + ∠B + ∠C = 1800
620 + 620 + ∠C = 1800
1240 + ∠C = 1800
∠C = 1800 – 1240
We get,
∠C = 560
Hence, ∠C = 560
6. In △ABC, ∠B = ∠C and ∠A = 1000; find ∠B.
Solution:
Given
∠B = ∠C
We know that,
Sum of all the angles in a triangle is 1800
∠A + ∠B + ∠C = 1800
1000 + ∠B + ∠B = 1800
1000 + 2∠B = 1800
2∠B = 1800 – 1000
We get,
2∠B = 800
∠B = 800 / 2
∠B = 400
Therefore, ∠B + ∠C = 400
7. Find, giving reasons, the unknown marked angles, in each triangle drawn below:
Solution:
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
So,
(i) 1100 = x + 300 [By property]
x = 1100 – 300
We get,
x = 800
(ii) x + 1150 = 1800 [By linear property of angles]
x = 1800 – 1150
We get,
x = 650
By exterior angle property
x + y = 1150
650 + y = 1150
y = 1150 – 650
We get,
y = 500
Therefore the value of angle x is 650 and y is 500
(iii) By exterior angle property,
1100 = 2x + 3x
5x = 1100
x = 1100 / 5
We get,
x = 220
Hence,
The value of 2x = 2 × 22
= 440
The value of 3x = 3 × 22
= 660
8. Classify the following triangles according to angle:
Solution:
(i) Since, one of the angle of a triangle is 1200.
Therefore, it is obtuse angled triangle
(ii) Since, all the angles of a triangle is less than 900
Therefore, it is acute angled triangle
(iii) Since ∠MNL = 900 and
Sum of two acute angle s,
∠M + ∠N = 300 + 600
= 900
Therefore, it is right angled triangle
9. Classify the following triangles according to sides:
(i)
(ii)
Solution:
(i) In the given triangle, we find two sides are equal.
Therefore, it is isosceles triangle
(ii) In the given triangle, all the three sides are unequal.
Therefore, it is scalene triangle
(iii) In the given triangle, all the three sides are unequal.
Therefore, it is scalene triangle
(iv) In the given triangle, all the three sides are equal.
Therefore, it is equilateral triangle
Exercise 26(B)
1. Construct triangle ABC, when:
AB = 6 cm, BC = 8 cm and AC = 4 cm
Solution:
Given
AB = 6 cm
BC = 8 cm
AC = 4 cm
Now,
Steps of Construction:
(i) Draw a line AB of length 6 cm
(ii) Using compasses, take B as centre, and draw an arc of 8 cm radius
(iii) Again, taking A as centre, draw another arc of 4 cm radius, which cuts the previous arc at point C
(iv) Now join AC and BC
The obtained triangle ABC is the required triangle.
2. AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm
Solution:
Given
AB = 3.5 cm
AC = 4.8 cm
BC = 5.2 cm
Steps of Construction:
(i) Draw a line AB of length 3.5 cm
(ii) With the help of compasses, taking B as centre, draw an arc of 5.2 cm radius
(iii) Again with A as centre, draw an arc of 4.8 radius
(iv) Now, join AC and BC
3. AB = BC = 5 cm and AC = 3 cm. Measure angles A and C. Is ∠A = ∠C?
Solution:
Given
AB = BC = 5 cm
AC = 3 cm
Steps of Construction:
(i) Draw a line AB of length 5 cm
(ii) Using compasses take B as centre and draw an arc of 5 cm radius
(iii) Now, taking A as centre, draw another arc of 3 cm radius, which cuts the previous arc at point C
(iv) Now, join AC and BC
Measuring angles A and C, we get
∠A = 72.5o and ∠C = 72.5o
Hence, yes ∠A = ∠C.
4. AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal?
Solution:
Given
AB = BC = CA = 4.5 cm
Steps of Construction:
(i) Draw a line AB of length 4.5 cm
(ii) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius
(iii) Again taking AC as centre, draw another arc of 4.5 cm radius, which cuts the previous arc at point C
(iv) Now, join AC and BC
(v) All the angles in ABC i.e ∠A = ∠B = ∠C = 600
Since AB = BC = CA = 4.5 cm and all the angles are equal. Hence, it is an equilateral triangle
5. AB = 3 cm, BC = 7 cm and ∠B = 900
Solution:
Given
AB = 3 cm,
BC = 7 cm and
∠B = 900
Steps of Construction:
(i) Draw a line segment AB of length 3 cm
(ii) Using compasses, construct ∠ABC = 900
(iii) Taking B as centre, draw an arc of 7 cm length and mark as point C i.e BC = 7 cm
(iv) Now, join A and C
(v) The obtained â–³ABC, is the required triangle
6. AC = 4.5 cm, BC = 6 cm and ∠C = 600
Solution:
Given
AC = 4.5 cm
BC = 6 cm
∠C = 600
Steps of Construction:
(i) Draw a line AC of length 4.5 cm
(ii) Using compasses, construct ∠ACB = 600
(iii) Draw an arc of 6 cm radius and mark it as B such that BC = 6 cm
(iv) Now, join B and A
7. AC = 6 cm, ∠A = 600 and ∠C = 450. Measure AB and BC.
Solution:
Given
AC = 6 cm
∠A = 600
∠C = 450
Steps of Construction:
(i) Draw a line segment AC of length 6 cm
(ii) With the help of Compass, construct ∠A = 600
(iii) Again, using compass, construct ∠C = 450
(iv) AD and CE intersect each other at point B
(v) Now, the obtained â–³ABC is the required triangle
(vi) Measure the side AB and BC with the help of a scale
(vii) We get, AB = 4.4 cm and BC = 5.4 cm
8. AB = 5.4 cm, ∠A = 300 and ∠B = 900. Measure ∠C and side BC.
Solution:
Given
AB = 5.4 cm
∠A = 300 and
∠B = 900
Steps of Construction:
(i) Draw a line segment AB of length 5.4 cm
(ii) With the help of compass, construct ∠A = 300
(iii) Similarly, construct ∠B = 900
(iv) AD and BE intersect each other at point C
(v) Hence, the obtained â–³ABC is the required triangle
(vi) On measuring we get, ∠C = 60 and side BC = 3.1 cm approximately
9. AB = 7 cm, ∠B = 1200 and ∠A = 300. Measure AC and BC.
Solution:
Given
AB = 7 cm
∠B = 1200
∠A = 300
Steps of Construction:
(i) Draw a line segment AB of length 7 cm
(ii) With the help of a compass, construct ∠A = 300
(iii) AE and BD intersect each other at point C
(iv) Hence, the obtained â–³ABC is the required triangle
(v) On measuring the lengths, we get AC = 12 cm and BC = 7 cm respectively
10. BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle
Solution:
Given
BC = 3 cm
AC = 4 cm and
AB = 5 cm
Steps of Construction:
(i) Draw a line segment AB of length 5 cm
(ii) From B, using compass cut an arc of 3 cm radius
(iii) Similarly, from A again with the help of compass, cut an arc of 4 cm bisecting the previous arc formed from point B
(iv) Now, join point C with A and B
(v) The obtained triangle is the required â–³ABC
(vi) On measuring ∠ACB, we get ∠ACB = 900.
Therefore, the obtained triangle ABC is a right angled triangle
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